No RQs for Wednesday. However, I wanted to reiterate the proof that I gave at the end of class today (addressing the RQ due for today’s class) and give you something to think about for when we get together again on Wednesday.

First, let’s prove:

Proposition: Let \(R\) be a commutative ring, \(a\in R\) a unit, and \(b\in R\) a zero divisor. Then \(ab\) is a zero divisor.

Proof. Since \(a\) is a unit, it is not a zero divisor. Thus, \(ab\neq 0\), since \(b\) being a zero divisor means that \(b\neq 0\) and the only solution to \(ax=0\) is \(x=0\) since \(a\) is a unit. Since \(b\) is a zero divisor, there is a \(c\neq 0\) in R such that \(bc=0\). Now consider \((ab)c\). This is the product of two nonzero elements of \(R\). Using associativity, we see that \((ab)c=a(bc)\). Since \(bc=0\), we thus have \((ab)c=a(bc)=a\cdot 0=0\). Thus, \(ab\) is a zero divisor with complementary zero divisor \(c\). ∎

Notice that our proof just barely used that \(a\) is a unit. In fact, all we needed was that \(a\) is not a zero divisor. Thus, we could revise the statement to be:

Proposition: Let \(R\) be a commutative ring, \(a\) a nonzero element of \(R\) that is not a zero divisor, and \(b\in R\) a zero divisor. Then \(ab\) is a zero divisor.

So here’s your question to think about for Wednesday: Suppose that \(R\) is a commutative ring and \(a,b\in R\) are units. Is \(ab\) always a unit? If so, what’s it’s multiplicative inverse? If not, find a commutative ring \(R\) and units \(a,b\in R\) such that \(ab\) is not a unit.